Simplify the following expression: $y = \dfrac{-6x^2+13x- 2}{-6x + 1}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-6)}{(-2)} &=& 12 \\ {a} + {b} &=& &=& {13} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $12$ and add them together. The factors that add up to ${13}$ will be your ${a}$ and ${b}$ When ${a}$ is ${1}$ and ${b}$ is ${12}$ $ \begin{eqnarray} {ab} &=& ({1})({12}) &=& 12 \\ {a} + {b} &=& {1} + {12} &=& 13 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-6}x^2 +{1}x) + ({12}x {-2}) $ Factor out the common factors: $ x(-6x + 1) - 2(-6x + 1)$ Now factor out $(-6x + 1)$ $ (-6x + 1)(x - 2)$ The original expression can therefore be written: $ \dfrac{(-6x + 1)(x - 2)}{-6x + 1}$ We are dividing by $-6x + 1$ , so $-6x + 1 \neq 0$ Therefore, $x \neq \frac{1}{6}$ This leaves us with $x - 2; x \neq \frac{1}{6}$.